What is the solution for y^{''}+4x^2+4y=6 ?

The solution for y^{''}+4x^2+4y=6 is y=c_{1}cos(2x)+c_{2}sin(2x)-x^2+2

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y^{''}+4x^2+4y=6

y^{^{''}} + 4 x^{2} + 4 y = 6

y = c_{1} cos (2 x) + c_{2} sin (2 x) - x^{2} + 2

Solution steps

y^{''} (x) + 4 x^{2} + 4 y = 6

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - x^{2} + 2

y = c_{1} cos (2 x) + c_{2} sin (2 x) - x^{2} + 2

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 8 y = x e^{x}

y^{^{''}} - 9 y^{^{'}} + 14 y = 12 - e^{x}

y^{^{''}} + 2 y^{^{'}} + 5 y = 45, y (0) = 2, y^{^{'}} (0) = 3

y^{^{''}} + y^{^{'}} - 20 y = e^{x} (82 + 84 x - 36 x^{2})

\frac{d ^{2} y}{d t ^{2}} + 6 \frac{d y}{d t} + 8 y = H (t)

What is the solution for y^{''}+4x^2+4y=6 ?
The solution for y^{''}+4x^2+4y=6 is y=c_{1}cos(2x)+c_{2}sin(2x)-x^2+2